How do you find the range of #f(x)=abs(x^2-8x+7)# for the domain #3

Question

How do you find the range of #f(x)=abs(x^2-8x+7)# for the domain #3 <= x <= 8#?

Isaiah Anthony · Accepted Answer

There are a few strategies that will be very effective for this. This is the one I employed: Graph the parabola #y = x^2-8x+7#, the flip the negative part across the #x# axis to get the absolute value. I chose this method, because it's relatively easy to find the #x# intercepts:
#x^2-8x+7 = 0# #(x-7)(x-1) = 0#, the intercepts are #1,7# and the parabola opens upward: The vertex has #x# coordinate equal to the midpoint between the intercepts, or #x=4# and #y# value 
#y=(4)^2-8(4)+7 = 16-32+7 = -11# graph{y = x^2-8x+7 [-18.32, 27.29, -11.85, 10.94]} The absolute values are not graphed: graph{y =abs(x^2-8x+7) [-20.61, -1.88, 12.35, -7.87]} Now:  #f(3) = 8#  and #f(8) = 7#, but between #3# and #8# we will hit f(7)=0, so the range runs from #0# up to somewhere.   The maximum value of the absolute value on that domain will be the absolute value of the minimum of the parabola or #11# The range for that domain is   #0 <= y <= 11#

Ethan Adkins · Answer

undefined To find the range of $ f(x) = |x^2 - 8x + 7| $ for the domain $ 3 \leq x \leq 8 $, evaluate the function at the endpoints of the domain and any critical points within that domain, and then determine the minimum and maximum values.

How do you find the range of #f(x)=abs(x^2-8x+7)# for the domain #3 &lt;= x &lt;= 8#?

How do you find the range of #f(x)=abs(x^2-8x+7)# for the domain #3 <= x <= 8#?